123246/132264*264396/246369这道数学题怎么做?

1

123246/132264*264396/246369
= [（132264+132132）/132264]/[(123246+123123)/123246]
=(1+132132/132246)/(1+123123/123246)
=[1+(132246-132)/132246]/[1+(123246-123)/123246]
=(1+1-1/1002)/(1+1-1/1002)
=1

知道是1 能做出的人很少要守信用给高分做

只能考虑用除以最大公约数的方法把分子分母尽量缩小来计算
用gcd()表示最大公约数，利用欧几里得辗转相除法:
gcd(123246,132264) = gcd(123264,9018) = gcd(9018,6030) = gcd(6030,2988) = gcd(2988,54) = gcd(54,18) = 18
123246/132264 = 6847 / 7348
gcd(264396,246369) = gcd(246396,18027) = gcd(18027,12045) = gcd(12045,5982) = gcd(5982,81) = gcd(81,69) = 3
2634396/246369 = 88132 / 82123
gcd(6847,82123) = gcd(6847,6806) = gcd(6806,41) = 1
gcd(88132,7348) = gcd(7348,7304) = gcd(7304,44) = gcd(44,0) = 44
原式 = 6847*2003/(82123*167)
6847/167 = 41
82123/2003 = 41
所以原式 = 1

更具体的一个解法，令123=a,分子为1002a*[1000*(2a+18)+3*(a+9)]，分母为2003a*[1000*(a+9)+2(a+9)]，约掉a.{1002*[1000*(2a+18)+3*(a+9)]}/2003*[1000*(a+9)+2(a+9)],.{1002*[2003a+18027}/{2003*[